Window Queries πŸ‘» - Google Top Interview Questions


Problem Statement :


You are given two lists of integers nums and queries. 

You are also given an integer w. Return a new list where for each query q in queries, we answer the question of how many number of different ways are there for a window of length w to cover an element with value q.

Constraints

n ≀ 100,000 where n is the length of nums

m ≀ n where m is the length of queries

0 ≀ nums[i] < n

0 ≀ queries[i] < n

Example 1

Input

nums = [2, 1, 2, 3, 4]

queries = [2, 1]

w = 3

Output

[3, 2]

Explanation

For the first query, we ask how many ways are there for a window of length 3 to contain the value 2. 
There are three ways:



[2, 1, 2]

[1, 2, 3]

[2, 3, 4]

For the second query, we ask how many ways are there for a window of length 3 to contain the value 1. 
There are two ways:



[2, 1, 2]

[1, 2, 3]

Example 2

Input

nums = [2, 2, 2]

queries = [2]

w = 2

Output

[2]

Explanation

The two ways are [2, 2] and [2, 2].


Solution :



title-img 




                        Solution in C++ :

int solve(vector<int>& locs, int w, int maxindex) {
    // for the given set of locations, how many distinct windows
    // of width w are attainable
    // if the maximum usable index is maxindex?
    vector<pair<int, int>> v;
    for (int i = 0; i < locs.size(); i++) {
        int lhs = max(0, locs[i] - w + 1);
        int rhs = min(maxindex, locs[i]);
        // a window starting at any index in [lhs, rhs] covers this location
        v.emplace_back(lhs, rhs);
    }
    int ret = 0;
    for (int i = 0; i < v.size(); i++) {
        if (i == 0) {
            ret += v[i].second - v[i].first + 1;
        } else {
            // check for overlap with the previous window
            int nlhs = max(v[i - 1].second + 1, v[i].first);
            ret += v[i].second - nlhs + 1;
        }
    }
    return ret;
}

vector<int> solve(vector<int>& nums, vector<int>& queries, int w) {
    unordered_map<int, vector<int>> locs;
    for (int i = 0; i < nums.size(); i++) {
        locs[nums[i]].push_back(i);
    }
    unordered_map<int, int> dp;
    for (auto out : locs) {
        dp[out.first] = solve(out.second, w, nums.size() - w);
    }
    vector<int> ret;
    for (int out : queries) ret.push_back(dp[out]);
    return ret;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int[] solve(int[] nums, int[] queries, int w) {
        int[] result = new int[queries.length];
        // Count element q[i] in range(start , start + w)
        if (nums == null || nums.length == 0)
            return result;
        Map<Integer, List<Integer>> idxMap = new HashMap<>();

        for (int i = 0; i < nums.length; i++) {
            List<Integer> idxList = idxMap.getOrDefault(nums[i], new ArrayList<Integer>());
            idxList.add(i);
            idxMap.put(nums[i], idxList);
        }

        for (int i = 0; i < queries.length; i++) {
            List<Integer> idxList = idxMap.get(queries[i]);
            int l = 0, r = 0, right = -1;
            for (int index : idxList) {
                l = Math.max(index - w + 1, right + 1);
                r = Math.min(index, nums.length - w);
                result[i] += r - l + 1;
                right = r;
            }
        }
        return result;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, A, queries, W):
        N = len(A)
        index = collections.defaultdict(list)
        for i, x in enumerate(A):
            index[x].append(i)

        def ways(q):
            ans = 0
            right = -1

            for i in index[q]:
                l = max(i - W + 1, right + 1)
                r = min(i, N - W)
                ans += r - l + 1
                right = r

            return ans

        return [ways(q) for q in queries]


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